<meta http-equiv="refresh" content="1; url=/nojavascript/"> Stoichiometric Calculations | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Chemistry - Intermediate Go to the latest version.

# 12.2: Stoichiometric Calculations

Created by: CK-12
0  0  0

## Lesson Objectives

• Calculate the amount in moles of a reactant or product from the mass of another reactant or product. Calculate the mass of a reactant or product from the moles of another reactant or product.
• Calculate the mass of a reactant or product from the mass of another reactant or product.
• Create volume ratios from a balanced chemical equation.
• Use volume ratios and other stoichiometric principles to solve problems involving mass, molar amounts, or volumes of gases.

### Recalling Prior Knowledge

• How can a balanced chemical equation be used to construct mole ratios between substances?
• How is a mole ratio used to convert from moles of one reactant or product to moles of another?

The mole ratio is the essence of ideal stoichiometry. The mole ratio tells us the quantitative relationship between reactants and products under ideal conditions, in which all reactants are completely converted into products. In the laboratory, most reactions are not completely ideal. Reactions may not proceed 100% to completion or a given set of reactants may undergo several reactions leading to different products. However, theoretical stoichiometric calculations are important because they allow chemists to know the maximum amount of product that can expect to be produced in a given reaction.

## Ideal Stoichiometry

Solving any stoichiometric calculation starts with a balanced chemical equation. As we saw in the last lesson, “Mole Ratios,” the coefficients of the balanced equation are the basis for the mole ratio between any pair of reactants and/or products. The following flowchart (Figure below) shows the conversion from a given substance in moles to moles of an unknown substance involves multiplying by the mole ratio.

Flowchart shows how a mole ratio is used in a stoichiometric conversion problem.

In this lesson, you will expand your understanding of stoichiometry to include the amounts of substances that are measured either by mass or by volume.

## Mass-Based Stoichiometry

While the mole ratio is ever-present in all stoichiometry calculations, amounts of substances in the laboratory are most often measured by mass. Therefore, we need to use mole-mass calculations in combination with mole ratios to solve several different types of mass-based stoichiometry problems.

### Mass to Moles Problems

In this type of problem, the mass of one substance is given, usually in grams. From this, you are to determine the amount in moles of another substance that will either react with or be produced from the given substance.

mass of given → moles of given → moles of unknown

The mass of the given substance is converted into moles by use of the molar mass of that substance from the periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation.

Sample Problem 12.3: Mass-Mole Stoichiometry

Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.

Sn(s) + 2HF(g) → SnF2(s) + H2(g)

How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?

Step 1: List the known quantities and plan the problem.

Known

• given: 75.0 g Sn
• molar mass of Sn = 118.69 g/mol
• 1 mol Sn = 2 mol HF (mole ratio)

Unknown

• mol HF

Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.

g Sn → mol Sn → mol HF

Step 2: Solve.

$75.0 \ \text{g Sn} \times \dfrac{1 \ \text{mol Sn}}{118.69 \ \text{g Sn}} \times \dfrac{2 \ \text{mol HF}}{1 \ \text{mol Sn}} = 1.26 \ \text{mol HF}$

The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.

Practice Problem
1. Silver oxide is used in small batteries called button batteries (Figure below). It decomposes upon heating to form silver metal and oxygen gas.
2Ag2O(s) → 4Ag(s) + O2(g)

How many moles of silver are produced by the decomposition of 4.85 g of Ag2O?

Silver oxide is used in button batteries such as these watch batteries.

### Moles to Mass Problems

In this type of problem, the amount of one substance is given in moles. From this, you are to determine the mass of another substance that will either react with or be produced from the given substance.

moles of given → moles of unknown → mass of unknown

The moles of the given substance are first converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Then, the moles of the unknown are converted into mass in grams by use of the molar mass of that substance from the periodic table.

Sample Problem 12.4: Mole-Mass Stoichiometry

Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor.

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)

What mass of oxygen gas is consumed in a reaction that produces 4.60 mol SO2?

Step 1: List the known quantities and plan the problem.

Known

• given: 4.60 mol SO2
• 2 mol SO2 = 3 mol O2 (mole ratio)
• molar mass of O2 = 32.00 g/mol

Unknown

• mass O2 = ? g

Use the mole ratio to convert from mol SO2 to mol O2. Then convert mol O2 to grams. This will be done in a single two-step calculation.

mol SO2 → mol O2 → g O2

Step 2: Solve.

$4.60 \ \text{mol SO}_2 \times \dfrac{3 \ \text{mol O}_2}{2 \ \text{mol SO}_2} \times \dfrac{32.00 \ \text{g O}_2}{1 \ \text{mol O}_2} = 221 \ \text{g O}_2$

According to the mole ratio, 6.90 mol O2 is produced, which has a mass of 221 g. The answer has three significant figures because the given moles has three significant figures.

Practice Problem
1. Copper metal reacts with sulfur to form copper(II) sulfide. What mass of copper(II) sulfide is produced by the reaction of 0.528 mol Cu with excess sulfur?

### Mass to Mass Problems

Mass-mass calculations are the most practical of all mass-based stoichiometry problems. Moles cannot be measured directly, while the mass of any substance can generally be easily measured in the lab. This type of problem is three steps and is a combination of the two previous types.

mass of given → moles of given → moles of unknown → mass of unknown

The mass of the given substance is converted into moles by use of the molar mass of that substance from the periodic table. Then, the moles of the given substance are converted into moles of the unknown by using the mole ratio from the balanced chemical equation. Finally, the moles of the unknown are converted to mass by use of its molar mass.

Sample Problem 12.5: Mass-Mass Stoichiometry

Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.

NH4NO3(s) → N2O(g) + 2H2O(l)

In a certain experiment, 45.7 g of ammonium nitrate is decomposed. Find the mass of each of the products formed.

Step 1: List the known quantities and plan the problem.

Known

• given: 45.7 g NH4NO3
• 1 mol NH4NO3 = 1 mol N2O = 2 mol H2O (mole ratios)
• molar mass of NH4NO3 = 80.06 g/mol
• molar mass of N2O = 44.02 g/mol
• molar mass of H2O = 18.02 g/mol

Unknown

• mass N2O = ? g
• mass H2O = ? g

Perform two separate three-step mass-mass calculations as shown below.

g NH4NO3 → mol NH4NO3 → mol N2O → g N2O
g NH4NO3 → mol NH4NO3 → mol H2O → g H2O

Step 2: Solve.

$45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{1 \ \text{mol N}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{44.02 \ \text{g N}_2\text{O}}{1 \ \text{mol N}_2\text{O}} = 25.1 \ \text{g N}_2\text{O}$
$45.7 \ \text{g NH}_4\text{NO}_3 \times \frac{1 \ \text{mol NH}_4\text{NO}_3}{80.06 \ \text{g NH}_4\text{NO}_3} \times \frac{2 \ \text{mol H}_2\text{O}}{1 \ \text{mol NH}_4\text{NO}_3} \times \frac{18.02 \ \text{g H}_2\text{O}}{1 \ \text{mol H}_2\text{O}} = 20.6 \ \text{g H}_2\text{O}$

The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures.

Practice Problem
1. Solid iron(III) hydroxide reacts with sulfuric acid to produce aqueous iron(III) sulfate and water. What mass of sulfuric acid is needed to completely react with 12.72 g of iron hydroxide? What mass of iron(III) sulfate is produced?

## Volume-Based Stoichiometry

As you learned in a previous chapter, Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain the same number of gas particles. Further, one mole of any gas at standard temperature and pressure (0°C and 1 atm) occupies a volume of 22.4 L. These characteristics makes stoichiometry involving gases at STP is very straightforward. Consider the reaction of nitrogen and oxygen cases to form nitrogen dioxide.

$& \text{N}_{2(g)} && + && 2\text{O}_{2(g)} && \rightarrow && 2\text{NO}_{2(g)} \\& 1 \ \text{molecule} && && 2 \ \text{molecules} && && 2 \ \text{molecules} \\& 1 \ \text{mol} && && 2 \ \text{mol} && && 2 \ \text{mol} \\& 1 \ \text{volume} && && 2 \ \text{volumes} && && 2 \ \text{volumes}$

Because of Avogadro’s realization, the mole ratio between substances in a gas-phase reaction are also volume ratios. The six possible volume ratios for the above equation are:

1. $\dfrac{1 \ \text{volume N}_2}{2 \ \text{volumes O}_2}$ or $\dfrac{2 \ \text{volumes O}_2}{1 \ \text{volume N}_2}$
2. $\dfrac{1 \ \text{volume N}_2}{2 \ \text{volumes NO}_2}$ or $\dfrac{2 \ \text{volumes NO}_2}{1 \ \text{volume N}_2}$
3. $\dfrac{2 \ \text{volumes O}_2}{2 \ \text{volumes NO}_2}$ or $\dfrac{2 \ \text{volumes NO}_2}{2 \ \text{volumes O}_2}$

### Volume to Volume Problems

The volume ratios above can easily be used when the volume of one gas in a reaction is known and you need to determine the volume of another gas that will either react with or be produced from the first gas. The pressure and temperature conditions of both gases need to be the same.

Sample Problem 12.6: Volume-Volume Stoichiometry

The combustion of propane gas produces carbon dioxide and water vapor (see Figure below).

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

A municipal propane tank in Austin, TX. The combustion of propane gas produces carbon dioxide and water vapor.

What volume of oxygen is required to completely combust 0.650 L of propane? What volume of carbon dioxide is produced in the reaction?

Step 1: List the known quantities and plan the problem.

Known

• given: 0.650 L C3H8
• 1 volume C3H8 = 5 volumes O2
• 1 volume C3H8 = 3 volumes CO2

Unknown

• volume O2 = ? L
• volume CO2 = ? L

Step 2: Solve.

$0.650 \ \text{L C}_3\text{H}_8 \times \frac{5 \ \text{L O}_2}{1 \ \text{L C}_3\text{H}_8} = 3.25 \ \text{L O}_2$
$0.650 \ \text{L C}_3\text{H}_8 \times \frac{3 \ \text{L CO}_2}{1 \ \text{L C}_3\text{H}_8} = 1.95 \ \text{L CO}_2$

Because the coefficients of the O2 and the CO2 are larger than that of the C3H8, the volumes for those two gases are greater. Note that total volume is not necessarily conserved in a reaction because moles are not necessarily conserved. In this reaction, 6 total volumes of reactants becomes 7 total volumes of products.

Practice Problem
1. Using the combustion of propane equation from Sample Problem 12.6, what volume of water vapor is produced in a reaction that produces 320. mL of carbon dioxide?

### Mass to Volume and Volume to Mass Problems

Chemical reactions frequently involve both solid substances whose mass can be measured as well as gases for which measuring the volume is more appropriate. Stoichiometry problems of this type are called either mass-volume or volume-mass problems.

mass of given → moles of given → moles of unknown → volume of unknown
volume of given → moles of given → moles of unknown → mass of unknown

Because both types of problems involve a conversion from either moles of gas to volume or vice-versa, we can use the molar volume of 22.4 L/mol provided that the conditions for the reaction are STP. In a later chapter, we will see how to solve this type of problem when the conditions are not standard.

Sample Problem 12.7: Mass-Volume Stoichiometry

Aluminum metal reacts rapidly with aqueous sulfuric acid to produce aqueous aluminum sulfate and hydrogen gas.

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

Determine the volume of hydrogen gas produced at STP when a 2.00 g piece of aluminum completely reacts.

Step 1: List the known quantities and plan the problem.

Known

• given: 2.00 g Al
• molar mass Al = 26.98 g/mol
• 2 mol Al = 3 mol H2

Unknown

• volume H2 = ?

The grams of aluminum will first be converted to moles. Then the mole ratio will be applied to convert to moles of hydrogen gas. Finally, the molar volume of a gas will be used to convert to liters of hydrogen.

g Al → mol Al → mol H2 → L H2

Step 2: Solve.

$2.00 \ \text{g Al} \times \frac{1 \ \text{mol Al}}{26.98 \ \text{g Al}} \times \frac{3 \ \text{mol H}_2}{2 \ \text{mol Al}} \times \frac{22.4 \ \text{L H}_2}{1 \ \text{mol H}_2} = 2.49 \ \text{L H}_2$

The volume result is in liters. For much smaller amounts, it may be convenient to convert to milliliters. The answer here has three significant figures. Because the molar volume is a measured quantity of 22.4 L/mol, three is the maximum number of significant figures for this type of problem.

Sample Problem 12.8: Volume-Mass Stoichiometry

Calcium oxide is used to remove sulfur dioxide generated in coal-burning power plants according to the following reaction.

2CaO(s) + 2SO2(g) + O2(g) → 2CaSO4(s)

What mass of calcium oxide is required to react completely with 1.2 × 103 L of sulfur dioxide?

Step 1: List the known quantities and plan the problem.

Known

• given: 1.2 × 103 L SO2
• 2 mol SO2 = 2 mol CaO
• molar mass CaO = 56.08 g/mol

Unknown

• mass CaO = ? g

The volume of SO2 will be converted to moles, followed by the mole ratio, and finally a conversion of moles of CaO to grams.

L SO2 → mol SO2 → mol Ca → g CaO

Step 2: Solve.

$1.4 \times 10^3 \ \text{L SO}_2 \times \frac{1 \ \text{mol SO}_2}{22.4 \ \text{L SO}_2} \times \frac{2 \ \text{mol CaO}}{2 \ \text{mol SO}_2} \times \frac{56.08 \ \text{g CaO}}{1 \ \text{mol CaO}} = 3.5 \times 10^3 \ \text{g CaO}$

The resultant mass could be reported as 3.5 kg, with two significant figures. Even though the 2:2 mole ratio does not mathematically affect the problem, it is still necessary for unit conversion.

Practice Problem
1. Sodium azide (NaN3) is a compound that is used in automobile air bags (Figure below). A collision triggers its rapid decomposition into sodium and nitrogen gas, with the gas filling the air bag.
1. The decomposition of 1.00 g of NaN3 produces what volume of N2 at STP?
2. What mass of NaN3 is required to produce 250. L of N2 at STP?

The rapid decomposition of sodium azide produces nitrogen gas, which fills an automotive air bag.

### Other Stoichiometry

Stoichiometric conversions all involve mole ratios between substances in a balanced chemical equation. Problems that involve mass and/or volume of a gas are very common and practical. However, a third “arm” of the mole road map could also be part of a problem – the number of representative particles of a substance. Using the equation in Sample Problem 12.7, we could determine the number of formula units of aluminum sulfate produced when 25.0 g of Al reacts.

$25.0 \ \text{g Al} \times \frac{1 \ \text{mol Al}}{26.98 \ \text{g Al}} \times \frac{1 \ \text{mol Al}_2(\text{SO}_4)_3}{2 \ \text{mol Al}} \times \frac{6.02 \times 10^{23} \ \text{form. units Al}_2(\text{SO}_4)_3}{1 \ \text{mol Al}_2(\text{SO}_4)_3} = 2.79 \times 10^{23} \ \text{form. units Al}_2(\text{SO}_4)_3$

So problems could arise involving any combination of mass, volume, and number of representative particles. However, since particles cannot actually be counted and stoichiometry is used most often for lab-based situations, problems involving the number of particles are seldom encountered.

## Summary of Stoichiometry

The flowchart in Figure below illustrates the types of stoichiometry problems that we have seen in this chapter and that you will most often need to solve. Conversion (b) is always present in any stoichiometry problem, while the use of the conversions represented by (a), (c), (e), and (d) depend on the specific type of problem. Conversion (f) is unique to a volume-volume problem.

Flowchart for solving many types of stoichiometry problems.

An example of a stoichiometry problem can be seen at http://www.khanacademy.org/science/physics/thermodynamics/v/stoichiometry-example-problem-1.

A second example of a stoichiometry problem can be seen at http://www.khanacademy.org/science/physics/thermodynamics/v/stoichiometry-example-problem-2.

## Lesson Summary

• All stoichiometry problems involve the use of a mole ratio to convert between moles of a given substance and moles of an unknown substance.
• In an ideal stoichiometry problem, the mass of any reactant or product can be calculated if the balanced equation and the mass of another reactant or product are known.
• Since the molar volume of any gas at STP is constant, gas volumes can also be used in stoichiometric calculations. This allows for multiple types of problems involving mass, volume, and amount in moles.

## Lesson Review Questions

### Reviewing Concepts

1. How many calculations steps are involved in each of the following stoichiometry problems between two substances A and B from a chemical reaction?
1. moles of A → mass of B
2. volume of A → volume of B
3. mass of B → mass of A
4. moles of B → volume of A
2. Why does a mass-mass stoichiometry problem require three steps, while a volume-volume stoichiometry problem only requires one step?

### Problems

1. The double-replacement reaction of solutions of calcium chloride with silver nitrate produces a solution of calcium nitrate and a precipitate of silver chloride.
1. How many moles of calcium chloride are needed to react completely with 2.28 mol of silver nitrate?
2. If 0.0623 mol CaCl2 reacts completely, how many grams of AgCl are produced?
3. In order to produce exactly 1.50 g AgCl, how many moles of each of the two reactants should be used?
2. Silicon dioxide reacts upon heating with carbon to produce silicon carbide (SiC) and carbon monoxide.
1. What mass of carbon is required to react completely with 15.70 g of SiO2?
2. When 152 g SiO2 reacts with excess carbon, what mass of SiC is produced?
3. If a certain reaction produced 42.2 g CO, what mass of carbon reacted?
3. Butane (C4H10) combusts according to the reaction: Assume no change in temperature or pressure for the following questions.
1. What volume of O2 is needed to combust 425 mL of butane?
2. What volume of butane is combusted when 729 L of CO2 is produced?
3. When 6.20 L of butane is combusted, what volumes of CO2 and H2O are produced?
4. Dissolving calcium carbonate in hydrochloric acid produces aqueous calcium chloride, carbon dioxide, and water. Assume the reaction takes place at STP.
1. What volume of CO2 is produced by the reaction of 9.58 g CaCO3?
2. If 67.1 L of CO2 is produced in a reaction, what mass of HCl reacted?
3. In a certain reaction, 0.812 mol HCl is used. What mass of CaCO3 would react? What volume of CO2 would be produced?
5. Nitroglycerin is a an explosive compound and decomposes into multiple gaseous products according to the reaction:
1. What mass of nitrogen gas at STP is produced when 0.314 g of nitroglycerin decomposes?
2. What volume of carbon dioxide at STP is produced in a reaction that also produces 2.25 mol O2?
3. What is the total volume of all gases produced at STP by the full decomposition of 10.0 g of nitroglycerin?
6. Given the reaction for the rusting of iron to iron(III) oxide:
1. If a reaction produces 46.2 g of Fe2O3, how many Fe atoms reacted?
2. What volume of O2 at STP is needed to fully react with 8.39 × 1024 atoms of iron?
3. The complete reaction of 0.916 mol Fe will produce how many formula units of Fe2O3?
7. Zinc reacts with hydrochloric acid according to the equation: In a certain experiment, a 3.77 g sample of impure zinc is reacted with excess hydrochloric acid. If 1.09 L of H2 gas is collected at STP, what was the percentage of zinc in the original sample? Assume that the impurities do not react with HCl.

## Points to Consider

This lesson dealt with ideal stoichiometry, where 100% of reactants was converted to products. In the real world, many chemical reactions do not proceed entirely in this way.

• How will we calculate the amount of products formed in a reaction when two or more reactants are combined in a ratio other than the mole ratio from the balanced equation?
• How can we express the extent to which a set of reactants is converted to products if it is less than 100%?

Jan 31, 2013

Sep 09, 2014